Question

Testing Claims About Proportions. In Exercises 9–32, test the given claim. Identify the null hypothesis, alternative hypothesis, test statistic, P-value, or critical value(s), then state the conclusion about the null hypothesis, as well as the final conclusion that addresses the original claim. Use the P-value method unless your instructor specifies otherwise. Use the normal distribution as an approximation to the binomial distribution, as described in Part 1 of this section.

Survey Return Rate In a study of cell phone use and brain hemispheric dominance, an Internet survey was e-mailed to 5000 subjects randomly selected from an online group involved with ears. 717 surveys were returned. Use a 0.01 significance level to test the claim that the return rate is less than 15%.

Short answer

Nullhypothesis: The proportion of e-mails that were returned is equal to 15%.

Alternativehypothesis: The proportion of e-mails that were returned is less than 15%.

Teststatistic: -1.307

Criticalvalue: -2.3263

P-value: 0.0956

The null hypothesis is failed to reject.

There is not enough evidence to support the claim that the return rate of surveys is less than 15%.

Step-by-step solution

Given information

Out of 5000 e-mails distributed, 717 surveys were returned.

Hypotheses

The null hypothesis is written as follows:

The proportion of surveys that were returned equals15%.

$H_0:p=0.15$

The alternative hypothesis is written as follows:

The proportion of surveys that were returned is less than 15%.

$H_1:p<0.15$

The test is left-tailed.

Sample size, sample proportion, and population proportion

The sample size is n=5000.

The sample proportion of surveys that were returnedis as follows:

$$\begin{gathered} \hat{p}=\frac{717}{5000} \\ =0.143 \end{gathered}$$

The population proportion of surveys that were returned is 0.15.

Test statistic

The value of the test statistic is computed below:

$$\begin{gathered} z=\frac{\hat{p}-p}{\sqrt{\frac{pq}{n}}} \\ =\frac{0.143-0.15}{\sqrt{\frac{0.15\left(1-0.15\right)}{5000}}} \\ =-1.307 \end{gathered}$$

Thus, z=-1.307.

Critical value and p-value

Referring to the standard normal table, the critical value of z at $\alpha =0.01$ for a left-tailed test is -2.3263.

Referring to the standard normal table, the p-value for the test statistic value of-1.307 equals0.0956.

Since the p-value is greater than 0.05, the null hypothesis is failed to reject.

Conclusion of the test

There is not enough evidence to support the claim that the proportion of surveys that were returned is less than 0.15.